STATIC COMBINED – MULTIAXIAL STRESSES

Part I

STRESS TRANSFORMATION & PRINCIPAL STRESSES

Stress at a Point in the three: x, y and z directionsIn a more general case, stress at a point will have 18

components of stress, corresponding to the 3 directions, x, y & z. There are 6 faces of a cube. On each face there are 3 stresses - one normal stress and two shear stresses. They are ±x, ±y, ±z , ±xy, ±yz , ±zx, ±yx, ±zy, and ±xz. Force equilibrium requires that the opposite components are equal and hence they reduce to 3 normal and 6 shear stresses, x, y, z , xy, yz , zx, yx, zy and xz.. Further moment equilibrium requires adjacent shear components (complementary shear stresses) are equal, reducing the unknown to 3 normal and 3 shear stresses.

They are x, y, z , xy, yz and zx

When all stress components except x (or y or z) are zero, it is uniaxial stress state in x (or y or z) direction. When only x, y and xy are present and the other 3 are zero, it is biaxial stress state in x & y directions.

Stress at a Point in x’, y’ and z’ directions The stresses at a point in x, y and z directions can be resolved in to

similar 6 components at the same point in a different sets of x’, y’ and z’ directions, which are inclined to the x, y and z coordinates by 1 , 2 and 3. This is called stress transformation. The new stress components will be different; but they are functions of x, y , z , xy , yz and zx and 1 , 2

and 3.

Principal Stresses at a PointWhen the angles 1, 2 and 3 are so chosen that all the three

shear components xy , yz and zx at a point become zero, then only the three normal stresses 1, 2 and 3 that exist, called principal stresses. This set of directions is called principal directions and the planes perpendicular to these directions are principal planes.

There are no shear stresses on the principal planes.

However these principal stresses are only functions of the 6 components of the stresses in the x, y and z directions and the corresponding 3 angles.

This also implies that the 6 components on a set of any 3 directions could be determined from the 3 principal stresses, if the corresponding 3

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angles are given. (Recall the Mohr’s Circle is all about this transformation. Don’t get scared now!)

Relation between principal and non-principal stressesSince there are 3(1, 2 & 3) principal stresses in a given set of 3

principal directions, the expression for the principal stress is a cubical equation incorporating 3 roots (1, 2 & 3), as a function of the 6 stress components (x, y , z , xy , yz and zx) in the 3 mutually perpendicular but non-principal directions(x, y & z).

3 - I12 + I2 – I3 = 0 (1)where, I1 = (x + y + z) (2)

I2 = [(x y + y z + z x ) – (xy2 + yz

2 + zx

2 )] (3)I3 = [(x y z + 2xy yz zx - (x yz

2 + yzx

2 + z xy2 )](4)

Invariants

The three Ii’s are called invariants. For a set of principal stresses, the invariants will be the same for transformation to any other set of directions, say x’, y’ and z’.

If so, it should equally hold good for the principal stresses also. This is very important for ease of application to problems. Therefore

I1 = (x + y + z) (2) = (x’ + y’ + z’) 2a) = (1 + 2 + 3) (Principal stresses here) (2b)

I2 = [(x y + y z + z x ) – (xy2

+ yz2 + zx

2 )] (3) = [(x’ y’ + y’ z’ + z’ x’ ) – (’xy

2 + ’yz

2 + ’zx

2 )] (3a) = (1 2 + 2 3 + 3 1 ) (Principal stresses here) (3b)

I3 = [(x y z + 2xy yz zx - (x yz2 + yzx

2 + z xy2

)] (4) = [(x’ y’ z’ + 2’xy ’yz ’zx - (x’ ’yz

2 + y’zx

2 + z ’xy2 )](4a)

= 1 2 3 (Principal stresses here) (4b)

For a biaxial stress state, (1) will be a quadratic equation, say in x-y directions and therefore all components containing z will be zero and I3 is zero. Then (1) reduces to

2 - I1 + I2 = 0 (5)

This will give two roots 1 and 2. This is the familiar Mohr’s circle equation.

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1, 2 = (I1/2) [(I1/2 )2 - I2] (6)

The students should check the validity of (6) to represent the familiar Mohr’s circle, by substituting values for I’s. 1, 2 = (x + y) /2 {[(x - y) /2]2 + xy

2} (6A)

Principal Stress-Strain RelationIt is known that strain in one direction affects the strain in the other

two perpendicular directions through the Poisson’s ratio. Therefore, the principal strains in the elastic range can be expressed as

1 = 1/E – (/E)(2+ 3) = [1 – (2+ 3)] /E (7) 2 = 2/E – (/E)(3+ 1) = [2 – (3+ 1)] /E (8)

3 = 3/E – (/E)(1+ 2) = [3 – (1+ 2)] /E (9)

Change in volumeConsider a cube of volume Vo = LWD The changed volume, after deformation V1= L1W1D1 = (L + δL)(W + δW)(D + δD) _

V1= L(1 + 1)W(1 + 2)D(1 + 3) = Vo(1 + 1)(1 + 2)(1 + 3) (10) Expanding and for small strains, omitting higher orders of i,s, V1= Vo[1 + (1+ 2+ 3)] (11)Change in volume ∆V = V1 – Vo = Vo(1+ 2+ 3)Volumetric strainV =∆V/ Vo = (1+ 2+ 3) (12)

Constant volume conditionIf volume is constant, before and after application of stress and deformation, ∆V/Vo = (1+ 2+ 3) = 0 (13)Add the three strains (7), (8) & (9) and equate to zero, 1+2 +3 = (1+2 +3)(1 – 2)/E = 0 (14)Since (1+2 +3) 0 in general, the only way is (1 – 2) = 0 or = 0.5 (15)For constancy of volume, therefore, = 0.5

STATIC COMBINED – MULTIAXIAL STRESSES

Part II

THEORIES OF STRENGTH / FAILURE

Criterion for strength or failure For most design purposes the simple uniaxial test results are used

as criteria for strength or failure.

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In a uniaxial test, the strength or failure is decided by either yield strength (Y) or fracture strength (f) as the case may be as desired by the design requirement and or as per agreement between the designer and the client.

This holds, when the loading that produces only uniaxial stress state. But when the loading that produces multi-axial stress state, a question arises as how to fix a criterion for strength or failure. The theories of failure answer this question.

Need for Theories of Failure

A theory of failure proposes a criterion for strength / failure under any combination of multi-axial stresses, so that the stress at failure from the uniaxial test results could be related to this combination and used to define failure condition.

Advantage of Theories of Failure

Theories of strength or failure provide certain specific criteria that enable the use of simple uniaxial test results to decide strength or failure under multi-axial stress state, thus avoiding expensive tests involving infinite combinations of multi-axial stresses.

Some familiar theories of failure are:

1 Normal stress theory (Rankine)2 Normal strain theory (St. Venant)3 Shear stress theory (Guest, Tresca, Coulomb)4 Strain energy theory (Beltrami)5 Distortion energy / Octahedral shear stress theory

(Maxwell, von Mises-Henky)6 Internal friction theory (Mohr)

The last theory - Internal friction theory is special for cases where the fracture strength in compression is different from that in tension; normally cast iron, concrete, rock and the like fall under this category. Internal friction theory is not considered here.

All the other theories assume the strength in tension and compression to be equal.

Here the theories are developed first in terms of the principal stresses. When only non-principal stresses are known, (2a), (3a), (4a) and (5)/(6) will be used to find / represent principal stresses, when found necessary.

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Failure Strength in Uniaxial and Multi-axial Stress State

The failure stress or strain in a uniaxial test may correspond either to yielding, ultimate strength or fracture. This will be decided / specified by the user/ organization/ client.

For getting an insight to these theories, we use here only yielding as failure condition. Where necessary, fracture could be substituted. A common statement for all the theories is given below. The student may choose the suitable phrase to define respectively each theory.

Yielding of the element subjected to combined stress-state occurs when the loading has reached such a value that the maximum

1) principal stress becomes equal to the stress at yielding in simple uniaxial tension/compression test.

2) principal strain becomes equal to the strain …3) shear stress becomes equal to the shear stress… 4) strain energy/volume equal to the strain energy/volume …5) distortion energy/volume becomes equal distortion

energy/volume …

1 Maximum Normal / Principal Stress Theory (Rankine) The failure under multi-axial stress condition occurs when the

principal stress that has the maximum magnitude, exceeds the uniaxial stress at yield Y.

In many cases, it will not be clear as to which principal stress has the highest magnitude. Hence it is better to consider all the three principal stresses and check.

1 Y or 2 Y or 3 Y (I)

It is important to note that we need to know the values of 1, 2 and 3

individually to solve this problem. For this we should solve the cubical equation (1) in the triaxial stress state. Fortunately most practical problems are in biaxial stress state, where one of 1, 2 and 3 = 0 and then we need to solve only the quadratic equation (6) – Mohr’s circle - to find any two principal stresses. The other is zero.

2 Maximum Normal / Principal Strain Theory (St Venant’s)The failure under multi-axial stress condition occurs when the

principal strain that has the maximum magnitude, exceeds the uniaxial strain at yield.

1 Y or 2 Y or 3 Y (16)Using the simple elastic stress-strain relation Y = Y / E, (16) reduces to

1 Y / E or 2 Y / E or 3 Y / E (17)

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Further using (7), (8) and (9), (17) can be expressed as

1 -(2+ 3) Y

or2 -(3+ 1) Y (II)or

3 -(1+ 2) Y

(The student should do the intermediate steps to get Eqn II)

As in the case of Normal Stress Theory, we need to know the individual values of principal stresses. The procedure outlined there can be applied here too.

3 Maximum Shear Stress Theory (Coulomb, Guest, Tresca)

The failure under multi-axial condition occurs when the principal shear stress that has the maximum magnitude, exceeds the uniaxial shear stress at yield.

Note that the maximum shear stress at yield in a uniaxial case is Y/2.

Similarly the magnitude of the principal- maximum shear stresses are: 1 - 2 / 2 or 2 - 3 / 2 or 3 - 1./ 2. Therefore,

1 - 2 / 2 Y/2 or 2 - 3 Y or 3 - 1 Y (III)

Note:1 - 2 etc is not shear stress. Only 1 -2 /2 = shear stress.As in the case of Normal Stress Theory, we need to know the individual values of principal stresses. The procedure outlined there can be applied here too. However, it can be seen that for a biaxial stress state with z = yz = zx = 3 = 0, from (6),

1 - 2 = 2 [(I1/2 )2 - I2] (18)

In general, it is set mathematically 1> 2 > 3.

Hence the absolute maximum relates to 3 - 1 Y (IIIa)

IMPORTANT NOTE: In the above 3 theories, 3 cases each need to be checked. Which value to choose among the 3?

(i) For shaft diameter, vessel thickness design, one that gives bigger diameter, bigger thickness must be chosen.

(ii) For pressure, force, moment etc as well as diameter of a pressure vessel, one that gives smaller pressure, force,

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moment etc as well as smaller diameter of a pressure vessel must be chosen. (For pressure vessel, greater the thickness, smaller is the stress. But greater the diameter, greater is the stress – It is different for shaft diameter.)

4 Maximum Strain Energy Theory (Beltrami)The failure occurs when the strain energy under multi-axial

condition exceeds the uniaxial strain energy at yield. Assume linear elastic behaviour.

The uniaxial stress energy /volume at yield

U = YY /2 = Y2 /2E (19)

The strain energy in multi-axial stress state/ volume

U = (11 + 22+ 33 ) / 2 (20)

Using (7), (8) and (9) and eliminating 1, 2 and 3 (20) leads to

U = [(12 +2

2 + 32) –2 (12

+ 23 + 31)/2E (21)

= [(12 +2

2 + 32) –2 I2)/2E (21a)

The student should do the intermediate steps to get Eqns( 21) and (21a)

Student to check, that (19) results from (21), when Y(uniaxial) = 1 and 2 = 3 = 0.

For a more convenient form, the first term in (21a) can be written as

(12 +2

2 + 32) = (1

+2 + 3)2 –2 (12 + 23 + 31) (22)

= I12 - 2 I2 (22a)

Using (2a), (3a) and (22a), it can be seen that (21a) becomes,

U = [I12 - 2(1 + ) I2] /2E (23)

Equating (19) and (15), for failure,

(12 +2

2 + 32) –2 (12

+ 23 + 31) Y2 (IV)

Equating (19) and (23), for failure, I1

2 - 2(1 + ) I2 Y2 (IVa)

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Equation (IVa) is simpler to use, since the individual values of the principal stresses are not required.

5 Maximum Distortion Energy/Octahedral Shear Stress Theory (Maxwell, von Mises-Henky)

The failure occurs when the distortion energy under multi-axial condition exceeds the distortion energy in the uniaxial stress at yield.

It is necessary to understand what distortion energy is.

It could be guessed that a cube subjected to pressure /stress equal in all the 3 sides, should remain a cube, of course, of changed side lengths. This is called hydrostatic or volumetric stress state.

But if the stresses are not equal on all sides but different, the cube will not remain a cube. It is then said to have distortion also.

DERIVATION: The students need not mug it up! However, they should know how it is done. They must believe it can be derived and NO magic is there!!

In the general case, we could consider a stress state to consist of two parts:

One having a specified amount of equal (principal) normal stresses as hydrostatic portion (whether compression or tension but all equal) that causes only change in volume of the cube. The equal hydrostatic components of the principal stresses and strains are called volumetric stress v and volumetric strain v.

The other having, in general, unequal stresses, causes only distortion of the cubical shape in to rhomboid with no volume change. The distortion components of stress and strain are denoted as d and d.

Therefore the three principal stresses could be expressed as sum of the two sets of principal stresses as

1 = d1 + v (24)2 = d2 + v (25)3 = d3 + v (26)

Adding all the three(1 + 2 + 3) = (d1 + d2 + d3) + 3 v(27)

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Now recall that the stresses (d1, d2, d3) do not cause any change in volume; that is constancy of volume in respect of distortion stress state.(See (13)in Part I). For small values of strains this means

(d1 + d2 + d3) = 0 (28)

Recall also, for constancy of volume in general, from (14)

1+2 +3 = (1+2 +3)(1 – 2)/E = 0 (14)

But in the present case only the distortion does no volume change. Therefore, substituting distortion related stresses and strains in (14) we get

(d1 + d2 + d3) = (d1 + d2 + d3)(1 - 2)/E = 0 (29)

Since there is volume change in the total stress state in the present case, (1 - 2) is not zero. Therefore,

(d1 + d2 + d3) = 0 (30)

Substituting (30) in (27), we get

v = (1 + 2 + 3)/3 (31)

The total strain energy, U is considered to be the sum of the strain energies due to these two components. Therefore

U = Uv + Ud or Ud = U – Uv (32)

Now U is already known from (15) or easily from (17). Uv needs to be found. The appropriate apportioning of the hydrostatic stress, which is of same magnitude in all the three directions, can be found.

By definition (in elastic range)Uv = (vv + vv+ vv ) / 2 = 3vv / 2 (33)

From (7), (8) and (9), noting stress is same in all directions = v,

v = (1 –2)v/ E (34)

From (33) and (34)

Uv = 3 (1 –2)v2

/2 E (35)

Using (31) in (35),

Uv = 3(1 –2)(1 +2 + 3)2

/(9) /2 E (36)

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OrUv = (1 –2) I1

2 /6 E = (1/3 – 2/3) I1

2 /2 E (37)

From (23) and (37) (students to work out)

Ud = U - Uv = [(1 – 1/3 + 2/3) I12 - 2(1 + ) I2] /2E

= [(2/3)(1+ ) I1

2 - 2(1 + ) I2] /2E

Ud = (1 + )(I12 - 3I2) /3E (38)

Ud (uniaxial): All except 1 become zero in (38). At yield 1 = Y. Then I1 = Y and I2 = 0. From (38), therefore

Ud (uniaxial) = (1 + )Y2 /3E (39)

Equating (38) and (39), for failure

I12 - 3I2 Y

2 (V)

Alternative Expression: (Know this, as it appears in all text books) When the principal stress values are substituted in (V), the left side

becomes, (students to use (22) to get this),

(1 +2 + 3) 2 – 3(12

+ 23 + 31)

= (12 +2

2 + 32) – (12

+ 23 + 31)

= [2(12 +2

2 + 32) – 2(12

+ 23 + 31)]/2

= [( 1 - 2) 2

+ (2 - 3)2 + (3 - 1)2 ] / 2 (40)

Using (40) in (V),

( 1 - 2)2 +(2 - 3)2

+ (3 - 1)2 2Y2 (Va)

This expression (Va) is another popular one, found in almost all textbooks. But it needs the knowledge of individual values of principal stresses, which may be difficult to find in triaxial stress state. Hence (V) may be the best and easy to use.

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Relative Scope and merits & demerits of different theories of failures

The maximum normal stress theory assumes that the highest principal stress alone is the governing value defining the failure. This ignores the influences of the other two normal stresses. It is mostly in agreement for brittle materials. However, for certain combinations of stresses, the agreement between the test- results and this theory is poor.The maximum normal strain theory has a good agreement with thick-walled tubes and hence is used in the design of guns.

The maximum shear stress theory has a fair agreement with the test results of ductile materials. It is a widely used theory by engineering societies. It requires that the yield strength in tension and compression are equal. When one of the principal stresses has opposite sign, it differs very much from the maximum normal stress theory.The distortion energy theory has the best agreement with test results as it takes into account all the three principal stresses together. It is interesting that the normal stress theory, the shear stress theory and the distortion energy theory have same effect, where one principal stress 3, is zero and the other two are equal in magnitude 1 = 2.

SUMMARY of Theories of Failure

3 - I12 + I2 – I3 = 0 (1)I1 = (x + y + z)

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= (1 + 2 + 3) (2a)I2 = [(x y + y z + z x ) – (xy

2 + yz

2 + zx

2 )] = (1 2 + 2 3 + 3 1 ) (3a)I3 = [(x y z + 2xy yz zx - (x yz

2 + yzx

2 + z xy2

)] = 1 2 3 (4a)

Normal Stress Theory (Rankine)1 Y or 2 Y or 3 Y (I)

We need to know 1, 2 and 3 individually to solve this problem.

Normal Strain Theory (St Venant)1 -(2+ 3) Y

or2 -(3+ 1) Y (II)

or3 -(1+ 2) Y

We need to know individual values of principal stresses

Shear Stress Theory (Coulomb, Guest, Tresca) 1 - 2 Y or2 - 3 Y (III) or 3 - 1. Y

We need to know individual values of principal stresses

Strain Energy Theory (Beltrami)(1

2 +22 + 3

2) –2 (12 + 23 + 31) Y2 (IV)

I12 - 2(1 + ) I2 Y

2 (IVa)We do not need to know individual values of principal stresses

Distortion Energy/Octahedral Shear Stress Theory (Maxwell, von Mises-Henky)I1

2 - 3I2 Y2 (V)

We do not need to know individual values of principal stresses

(1 - 2)2 +(2 - 3)2

+ (3 - 1)2 2Y2 (Va)

We need to know individual values of principal stresses

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